2(x^2-4x+1)=13

Solution for 2(x^2-4x+1)=13 equation:

2(x^2-4x+1)=13

We move all terms to the left:

2(x^2-4x+1)-(13)=0

We multiply parentheses

2x^2-8x+2-13=0

We add all the numbers together, and all the variables

2x^2-8x-11=0

a = 2; b = -8; c = -11;
Δ = b2-4ac
Δ = -82-4·2·(-11)
Δ = 152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{38}}{2*2}=\frac{8-2\sqrt{38}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{38}}{2*2}=\frac{8+2\sqrt{38}}{4}
$